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\title{Basic Mathematics Solutions}
\author{Michael Rocke}

\begin{document}

\maketitle

\section{Chapter 1}

\subsection{Exercise §2}

Using Commutativity and Associativity in proving the following
\subsubsection{1}
$ (a+b) + (c+d) = (a+d) + (b+c) $ \\
$ a + b + c + d $ by associativity \\
$ a + b + c + d = a + b + d + c = a + d + b + c $ by commutativity\\

\subsubsection{2}
$ (a + b) + (c + d) = (a + c) + (b + d) $ \\
$ a + b + c + d $ by associativity \\
$ a + b + c + d  = a + c + b + d $ by commutativity

\subsubsection{14} Solve for x

$ -2 + x = 4 $ \\
$ x = 4 + 2  = 6$ 

\subsubsection{15}
$ 2 - x = 5 $\\
$ -x = 5 - 2 = 3$\\
$ x = -3 $\\

\subsubsection{16}
$ x - 3 = 7 $ \\
$ x = 7 + 3 = 10 $

\subsubsection{17}
$ -x + 4 = -1 $ \\
$ -x = -1 -4 = -5 $ \\
$ x = 5 $

\subsubsection{18}
$ 4 - x = 8 $ \\
$ -x = 8 - 4  = 4 $ \\
$ x = -4 $


\subsubsection{19}
$ -5 -x = -2 $ \\
$ -x = -2 + 5 = 3 $ \\
$ x = -3 $

\subsubsection{20}
$ -7 + x = - 10 $ \\
$ x = - 10 + 7 = -3 $

\subsubsection{21}
$ -3 + x = 4 $ \\
$ x = 4 + 3 = 7 $


\end{document}
